Optimal State Regulator through the Matrix Ricatti Equation

1. Optimal State Regulator
through the Matrix Riccati
Equation
Submitted to
Mrs. Shimi SL
Assistant Professor
Submitted By
Mr. Dhruv Upadhaya
162510
ME [ I&C ], Regular
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2. What we are finding
• Solution for Optimal Control problem for a linear multivariable
system with the quadratic criterion function
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3. Completely controllable plant
𝑥 = 𝐴𝑥 + 𝐵𝑢
Where 𝑥 is the n x 1 state vector
u is the p x 1 state vector
A is the n x n state vector
B is the n x p state vector
And the null state 𝒙 = 0 is the desired steady state.
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………..(1.1)
Mr. Dhruv Upadhaya

4. Find the control law
u = -K x(t)
Where K is p x n real constant unconstrained gain matrix,
that minimizes the following performance index subject to the initial
conditions x(0) ≜ x0;
J =
1
2 0
∞
(x 𝑇 𝑄x+uT Ru) dt
Where Q is n x n positive definite, real, symmetric, constant matrix, and
R is p x p positive definite, real, symmetric, constant matrix.
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………..(1.2)
………..(1.3)
Mr. Dhruv Upadhaya

5. There are several ways to solve this optimal control problem. We will
use the Lyapunov function approach.
Substituting the equation (1.1) in (1.2), we obtain
ẋ = A x – BK x = (A – BK)x
Since the (A, B) pair is completely controllable, there exists a feedback
matrix K such that the (A – Bk) is a stable matrix.
The controllability of the given plant, thus, ensures the existence of a K
that minimizes J.
Now we obtain the following design equations for the optimal control
problem.
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………..(1.4)
Mr. Dhruv Upadhaya

6. The time derivative of the Lyapunov function is
𝑉(𝑥) =
1
2
x 𝑇(𝑄 + KT RK)x
The Lyapunov function
V(x) =
1
2
x 𝑇 P x
Where P is a positive definite, real, symmetric, constant matrix, and
(A – BK)TP + P(A – BK) + KTRK + Q = 0
The performance index
J =
1
2
x 𝑇(0) P x(0)
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………..(1.5)
………..(1.6)
………..(1.7)
………..(1.8)
Mr. Dhruv Upadhaya

7. Since feedback matrix K is unconstrained, the optimum value of J is
independent of initial conditions. The optimal kij’s are obtained from the
equations
𝜕𝑃
𝜕kij
= 0
For all i, j.
Since R has been assumed to be a positive definite matrix, we can write
R = ΓT Γ
Where Γ is a non singular matrix. Then equation (1.7) can be written as
(AT – KTBT) P + P(A – BK) + KT ΓT ΓK + Q = 0
Which can be written as
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………..(1.9)
Mr. Dhruv Upadhaya

8. ATP + PA – [ΓK – (ΓT)-1BTP]T [ΓK – (ΓT)-1BTP] – PBR-1BTP + Q= 0
The condition 1.9 for the unconstrained minimization of J leads to the
following equations :
𝜕
𝜕kij
[(ΓK – (ΓT)−1BTP)T (ΓK – (ΓT)−1BTP)] = 0
Since the matrix within brackets is non negative definite, the minimum
occurs when it is zero, or when
ΓK = (ΓT)−1BTP
Hence
K = Γ−1(ΓT)−1BTP = R−1BTP
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………..(1.10)
………..(1.11)
Mr. Dhruv Upadhaya

9. Equation (1.11) gives the optimal gain matrix K. Thus the optimal
control law is given by
u t = –Kx t = −R−1BTPx(t)
The matrix P in equation (1.12) must satisfy equation (1.10), or the
following reduced equation:
ATP + PA – PBR−1BTP + Q = 0
This equation is called Riccati equation
We have assumed Q to be positive definite matrix. This makes 𝑉 𝑥 in
equation (1.5) always negative definite; therefore the optimal feedback
system asymptotically stable. Controllability of the (A, B) pair and
positive definiteness of Q are, thus, sufficient for the existence of
asymptotically stable optimal solution to the control problem.
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………..(1.12)
………..(1.13)
Mr. Dhruv Upadhaya

10. 1. Solve the matrix Ricatti equation (1.13) for positive definite matrix
P, and
2. Substitute this matrix P into equation (1.12); the resulting equation
gives optimal control law.
Once the designer has specified Q and R, representing his assessment of
the relative importance of the various terms in the performance index
the solution of 1.13 specifies the optimal control law 1.12. This yields
the optimal closed loop system.
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Design Steps
Mr. Dhruv Upadhaya

11. If the resulting transient response is unsatisfactory, the designer may
alter the weighting matrices Q and R and try again.
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12. 1. The matrix R has been assumed to be positive definite. This is a necessary
condition for the existence of the optimal solution to the control problem as seen
from Eqn. (1.12)
2. We have assumed that the plant (1.1) is completely controllable, and the matrix
Q in performance index J given by Eqn. (1.3) is positive definite. These are
sufficient conditions for the existence of asymptotically stable optimal solution
to the control problem. The requirement on matrix Q may be relaxed to a
positive semidefinite matrix with the pair (A, H) completely observable, where
Q = HTH.
3. The solution of Eqn. (1.13) is not unique. Of the several possible solutions, the
desired answer is obtained by enforcing the requirement that P be positive
definite. The positive definite solution of Eqn. (1.13) is unique.
4. In very simple cases, the Riccati equation can be solved analytically, but usually
a numerical solution is required. A number of computer programs for the
purpose are available.
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Comments:
Mr. Dhruv Upadhaya

13. Comments
1. It is important to be able to find out whether the sought-after
solution exists or not before we start working out the solution. This
is possible only if necessary conditions for the existence of
asymptotically stable optimal solution are established. A discussion
of this subject entails not only controllability and observability but
also the concepts of stabilizability and detectability.
2. Equation (1.13) is a set of n2 nonlinear algebraic equations. Since P
is a symmetric matrix, we need to solve only n (n + 1)/2 equations.
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